Answer:
The maximum possible daily profit is $11,500.
Step-by-step explanation:
Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
[tex]f(x) = ax^{2} + bx + c[/tex]
It's vertex is the point [tex](x_{v}, f(x_{v})[/tex]
In which
[tex]x_{v} = -\frac{b}{2a}[/tex]
If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]f(x_{v})[/tex]
In this question:
The maximum daily profit happens when [tex]x_{v}[/tex] cars are sold. This profit is [tex]P(x_{v})[/tex]
[tex]P(x) = -35x^{2} + 2100x - 20000[/tex]
So [tex]a = -35, b = 2100[/tex]
[tex]x_{v} = -\frac{2100}{2*(-35)} = 30[/tex]
The maximum possible daily profit is:
[tex]P(30) = -35*30^2 + 2100*30 - 20000 = 11500[/tex]
The maximum possible daily profit is $11,500.
Suppose we take repeated random samples of 50 college students from the same population and determine a 95% confidence interval for the mean GPA from each sample. Which of the following statements is true regarding the confidence intervals?
A. The intervals are centered around the population mean GPA.
B. The intervals are centered around the sample mean GPA.
C. 95% of the intervals will contain the sample mean in the long run.
D. 95% of the intervals will contain the population mean in the long run.
Answer:
B. The intervals are centered around the sample mean GPA.
D. 95% of the intervals will contain the population mean in the long run.
Step-by-step explanation:
Confidence interval:
Depends on two things: The sample mean and the margin of error.
Lower end: Sample mean - margin of error
Upper end: Sample mean + margin of error
This means that the intervals are centered around the sample mean.
x% level:
x% of the intervals will contain the population mean in the long run.
So the true statements are:
B. The intervals are centered around the sample mean GPA.
D. 95% of the intervals will contain the population mean in the long run.
A tire manufacturer wants to estimate the average number of miles that may be driven in a tire of a certain type before the tire wears out. Assume the population is normally distributed. A random sample of tires is chosen and are driven until they wear out and the number of thousands of miles is recorded, find the 97% confidence interval using the sample data.
Answer:
97% confidence interval for the average number of miles that may be driven is [26.78 miles, 33.72 miles].
Step-by-step explanation:
We are given that a random sample of tires is chosen and are driven until they wear out and the number of thousands of miles is recorded;
32, 33, 28, 37, 29, 30, 22, 35, 23, 28, 30, 36.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average number of miles = [tex]\frac{\sum X}{n}[/tex] = 30.25
s = sample standard deviation = [tex]\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }[/tex] = 4.71
n = sample of tires = 12
[tex]\mu[/tex] = population average number of miles
Here for constructing a 97% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.
So, 97% confidence interval for the population mean, [tex]\mu[/tex] is ;
P(-2.55 < [tex]t_1_1[/tex] < 2.55) = 0.97 {As the critical value of t at 11 degrees of
freedom are -2.55 & 2.55 with P = 1.5%}
P(-2.55 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.55) = 0.97
P( [tex]-2.55 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]2.55 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.97
P( [tex]\bar X-2.55 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+2.55 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.97
97% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-2.55 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+2.55 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]30.25-2.55 \times {\frac{4.71}{\sqrt{12} } }[/tex] , [tex]30.25+2.55 \times {\frac{4.71}{\sqrt{12} } }[/tex] ]
= [26.78 miles, 33.72 miles]
Therefore, 97% confidence interval for the average number of miles that may be driven is [26.78 miles, 33.72 miles].
-3/2 (x-1/3)>1/5-5x
What is the solution for this ^ inequality.
Answer:
[tex]x>\frac{-3}{35}[/tex]I hope this help
Step-by-step explanation:
If AD=BD, which of the following relationships can be proved and why?
B
o
A. A ACD= A BCD, because of ASA.
B. XACD N BOD because of SAS
C. There is not enough information to prove a relationship.
(D. A ACD S ABCD, because of AS
SUBMIT
< PREVIOUS
Answer: SAS
Step-by-step explanation:
Arc CD is Two-thirds of the circumference of a circle. What is the radian measure of the central angle? StartFraction 2 pi Over 3 EndFraction radians StartFraction 3 pi Over 4 EndFraction radians StartFraction 4 pi Over 3 EndFraction radians StartFraction 3 pi Over 2 EndFraction radians CAN YOU SHOW WORK PLEASE
Answer:
[tex]\theta = \frac{4\pi}{3}[/tex]
Step-by-step explanation:
Given
Let A represent the Length of Arc CD and C, represents the circumference
[tex]A = \frac{2}{3} C[/tex]
Required
Find the central angle (in radians)
The length of arc CD in radians is as follows;
[tex]A = r\theta[/tex]
Where r is the radius and [tex]\theta[/tex] is the measure of central angle
The circumference of a circle is calculated as thus;
[tex]C = 2\pi r[/tex]
From the question, it was stated that the arc length is 2-3rd of the circumference;
This means that
[tex]A = \frac{2}{3} C[/tex]
Substitute [tex]2\pi r[/tex] for C and [tex]r\theta[/tex] for A
[tex]A = \frac{2}{3} C[/tex] becomes
[tex]r\theta = \frac{2}{3} * 2\pi r[/tex]
[tex]r\theta = \frac{4\pi r}{3}[/tex]
Divide both sides by r
[tex]\frac{r\theta}{r} = \frac{4\pi r}{3}/r[/tex]
[tex]\frac{r\theta}{r} = \frac{4\pi r}{3} * \frac{1}{r}[/tex]
[tex]\theta = \frac{4\pi r}{3} * \frac{1}{r}[/tex]
[tex]\theta = \frac{4\pi}{3}[/tex]
Hence, the measure of the central angle; [tex]\theta = \frac{4\pi}{3}[/tex]
Answer:
The answer is C on Edge 2020
Step-by-step explanation:
I did the assignment
In Denver, Colorado, they experience a lot of snow in the winter. During the last
snow storm, it snowed for 3 straight days and the snow consistently accumulated at
a rate of inch per hour. How much snow did Denver get over three days?
Your answer
Answer:
Denver got 72 inches of snow over three days.
Step-by-step explanation:
Since it has snowed consistently for 3 days, accumulating an inch of snow per hour, over that number of days at least 72 inches of snow would have accumulated.
This is so because, since each day has 24 hours, in the event of a 3-day snowfall, it would have lasted 72 hours. Thus, while every hour a new inch of snow would accumulate, at the end of the storm the city of Denver would have accumulated 72 inches of snow (1 x 24 x 3 = 72).
there are only red counters and blue counters in a bag. Jim takes at random a counter from a bag. the probability that the counter is red is 0.45 Jim puts the counter back into the bag. Molly takes at random a counter from the bag. She puts the counter back in the bag. What is the probability that Jim and Molly take counters of different colours? Give your answer as a decimal
Answer:
0.495 probability that Jim and Molly take counters of different colours
Step-by-step explanation:
For each trial, there are only two possible outcomes. Either a blue counter is picked, or a red counter is picked. The counter is put back in the bag after it is taken, which means that we can use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The probability that the counter is red is 0.45
This means that [tex]p = 0.45[/tex]
Jim taken a counter, then Molly:
Two trials, so [tex]n = 2[/tex]
What is the probability that Jim and Molly take counters of different colours?
One red and one blue. So this is P(X = 1).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{2,1}.(0.45)^{1}.(0.55)^{1} = 0.495[/tex]
0.495 probability that Jim and Molly take counters of different colours
Which expression is equivalent to (–2y – x) – (5y – 9x)?
Answer:
8x-7y
Step-by-step explanation:
(–2y – x) – (5y – 9x)
Distribute the minus sign
-2y -x -5y +9x
Combine like terms
-x +9x - 2y -5y
8x-7y
Answer:
[tex]8x-7[/tex]
Step-by-step explanation:
[tex](-2y - x) - (5y - 9x)[/tex]
Distribute the negative sign.
[tex](-2y - x)-5y+9x[/tex]
[tex]-2y - x-5y+9x[/tex]
[tex]-2y -5y+9x- x[/tex]
Add or subtract like terms.
[tex]-7y+8x[/tex]
As director of the employee wellness and productivity program in your company, you are interested in comparing the effects of strength training, aerobic training, and yoga on decreasing rates of injury and absenteeism. The company has 9 divisions with roughly the same number of employees, and you randomly assign 3 divisions to participate in strength training, 3 to aerobic training, and 3 to yoga. Your alternative hypothesis is
Answer:
[tex]\mu_1 \neq \mu_2 \neq \mu_3[/tex]
Where [tex]\mu_1[/tex] is the average effect of strength training
[tex]\mu_2[/tex] is the average effect of aerobic training
[tex]\mu_3[/tex] is the average effect of yoga
Step-by-step explanation:
The aim of this study is to confirm whether the strength training, aerobic training, and yoga have equal effect on the decreasing rates of injury and absenteeism or not. The null hypothesis suggests that these three training have equal effect on the decreasing rates of injury and absenteeism because according to the null hypothesis, there is no statistical difference between observed variables.
The alternative hypothesis on the other hand suggests a statistical difference between the observed variables. In this case, the alternative hypothesis suggests that the observed variables have different effects on the decreasing rates of injury and absenteeism.
My alternative hypothesis as the director of the employee wellness and productivity program is [tex]\mu_1 \neq \mu_2 \neq \mu_3[/tex]
Where [tex]\mu_1[/tex] is the average effect of strength training
[tex]\mu_2[/tex] is the average effect of aerobic training
[tex]\mu_3[/tex] is the average effect of yoga
The alternative hypothesis is, [tex]\mu_1 \neq \mu_2 \neq \mu_ 3[/tex].
Where, [tex]\mu_1[/tex] is the average effect of strength training,
[tex]\mu_2[/tex] is the average effect of aerobic training.
[tex]\mu_3[/tex] is the average effect of yoga.
Given that,
As director of the employee wellness and productivity program in your company,
you are interested in comparing the effects of strength training, aerobic training, and yoga on decreasing rates of injury and absenteeism.
The company has 9 divisions with roughly the same number of employees, and you randomly assign 3 divisions to participate in strength training, 3 to aerobic training, and 3 to yoga.
We have to determine,
Your alternative hypothesis is.
According to the question,
The effects of strength training, aerobic training, and yoga on decreasing rates of injury and absenteeism.
The aim of this study is to confirm whether strength training, aerobic training, and yoga have equal effects on decreasing rates of injury and absenteeism or not.
The null hypothesis suggests that these three pieces of training have an equal effect on the decreasing rates of injury and absenteeism because according to the null hypothesis,
There is no statistical difference between observed variables.
The alternative hypothesis on the other hand suggests a statistical difference between the observed variables.
In this case, the alternative hypothesis suggests that the observed variables have different effects on the decreasing rates of injury and absenteeism.
The company has 9 divisions with roughly the same number of employees, and you randomly assign 3 divisions to participate in strength training, 3 to aerobic training, and 3 to yoga.
Therefore, The alternative hypothesis as the director of the employee wellness and productivity program is,
Where [tex]\mu_1[/tex] is the average effect of strength training,
[tex]\mu_2[/tex] is the average effect of aerobic training.
[tex]\mu_3[/tex] is the average effect of yoga.
To know more about the Hypothesis click the link given below.
https://brainly.com/question/23056080
are the following figures similar?
Answer:
Third option.
Step-by-step explanation:
The sides have to be proportional.
5/3 = 25/15
1.67 = 1.67
Yes; the corresponding sides are proportional.
Answer:
yes the corresponding sides are proportional
Step-by-step explanation:
figure ABCD has it length to be 5cm while width is 25cm
and figure EFGH has it length 3cm while the width is 15
Please answer this correctly
Answer:
54
Step-by-step explanation:
The pink parts are 9 out of total 11 parts.
9/11
Multiply with 66.
9/11 × 66
= 54
Hey there! :)
Answer:
P(Pink) = 54.
Step-by-step explanation:
Begin by calculating the possibility of the spinner landing on pink:
[tex]P(pink) = \frac{pink}{total}[/tex]
Therefore:
[tex]P(Pink) = \frac{9}{11}[/tex]
In this question, the spinner was spun 66 times. Since we have solved for the probability, we can set up ratios to find the probability of the spinner landing on pink out of 66.
[tex]\frac{9}{11}= \frac{x}{66}[/tex]
Cross multiply:
594 = 11x
Divide both sides by 11:
x = 54.
P(Pink) = 54.
Hayden picked 2/3 pound of blueberries. He then ate 1/3 of what he picked. How many pounds of blueberries did Hayden eat?
Answer:
2/3 x 1/3 = 2/9
2/9 lbs
Hope this helps
Step-by-step explanation:
Which system of inequalities is graphed below?
-5
Evaluate for f=3. 2f - f +7
Select the correct answer. Which of the following is equal to g(x)? (View the image)
answer A
Step-by-step explanation:
answer A
as g(0)=1/4
for the other cases g(0) = -1, 3 or 4 so this is not possible
Answer:
a:1/4x^2
Step-by-step explanation:
Leah has 28 more marbles than Dan. 1/3 of Leah’s marbles are equal in number to 4/5 of Dan’s marbles. Find the number of marbles Leah has.
Answer:
48
Step-by-step explanation:
L = D + 28
⅓L = ⅘D
Solve the system of equations using elimination or substitution. Using substitution:
⅓L = ⅘(L − 28)
Multiply both sides by 15:
5L = 12(L − 28)
Distribute:
5L = 12L − 336
Combine like terms:
336 = 7L
Divide:
L = 48
A consumer group was interested in comparing the operating time of cordless toothbrushes manufactured by two different companies. Group members took a random sample of 18 toothbrushes from Company A and 15 from Company B. Each was charged overnight and the number of hours of use before needing to be recharged was recorded. Company A toothbrushes operated for an average of 119.7 hours with a standard deviation of 1.74 hours; Company B toothbrushes operated for an average of 120.6 hours with a standard deviation of 1.72 hours. The 90% confidence interval is (-1.93, 0.13). The correct interpretation is:__________
A. We are 90% confident that, on average, there is no difference in operating hours between toothbrushes from Company A compared to those from Company B.
B. We are 90% confident that, on average, there is a difference in operating hours between toothbrushes from Company A compared to those from Company B.
C. We are 90% confident that, on average, the toothbrushes from Company B operate longer before needing to be recharged than the toothbrushes from Company A.
D. We are 90% confident that, on average, the toothbrushes from Company A operate longer before needing to be recharged than the toothbrushes from Company B.
Answer:
The degrees of freedom are given by:
[tex] df = n_A +n_B -2 = 18 +15-2= 31[/tex]
And the 90% confidence interval for this case is:
[tex] -1.90 \leq \mu_A -\mu_B \leq 0.13[/tex]
And for this case since the confidence interval contains the value 0 we can conclude that:
A. We are 90% confident that, on average, there is no difference in operating hours between toothbrushes from Company A compared to those from Company B.
Step-by-step explanation:
We know the following info given:
[tex] \bar X_A= 119.7[/tex] sample mean for A
[tex] s_A = 1.74[/tex] sample deviation for A
[tex] n_A = 18[/tex] sample size from A
[tex] \bar X_B= 120.6[/tex] sample mean for B
[tex] s_B = 1.72[/tex] sample deviation for B
[tex] n_B = 15[/tex] sample size from B
The degrees of freedom are given by:
[tex] df = n_A +n_B -2 = 18 +15-2= 31[/tex]
And the 90% confidence interval for this case is:
[tex] -1.90 \leq \mu_A -\mu_B \leq 0.13[/tex]
And for this case since the confidence interval contains the value 0 we can conclude that:
A. We are 90% confident that, on average, there is no difference in operating hours between toothbrushes from Company A compared to those from Company B.
The access code for a garage door consists of three digits. Each digit can be any number from 1 through 5, and each digit can be repeated. Complete parts (a) through (c). (a) Find the number of possible access codes. (b) What is the probability of randomly selecting the correct access code on the first try? (c) What is the probability of not selecting the correct access code on the first try? (a) Find the number of possible access codes. The number of different codes available is nothing.
Answer:
(a) 125
[tex](b) \dfrac{1}{125}[/tex]
[tex](c) \dfrac{124}{125}[/tex]
Step-by-step explanation:
We are given that access code consists of 3 digits.
Each digit can be any digit through 1 to 5 and can be repeated.
Now, this problem is equivalent to the problem that we have to find:
The number of 3 digit numbers that can be formed using the digits 1 to 5 with repetition allowed.
(a) We have 3 places here, unit's, ten's and hundred's places respectively and each of the 3 places have 5 possibilities (any digit allowed with repetition).
So, total number of access codes possible:
[tex]5\times 5 \times 5 = 125[/tex]
(b) Suppose, an access code is randomly selected, what is the probability that it will be correct.
Formula for probability of an event E can be observed as:
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}[/tex]
Here, only 1 code is correct, so
Number of favorable cases = 1
Total number of cases = 125
So, required probability:
[tex]P(E) = \dfrac{1}{125}[/tex]
(c) Probability of not selecting the correct access code on first time:
[tex]P(\overline E) = 1-P(E)\\\Rightarrow P(\overline E) = 1-\dfrac{1}{125}\\\Rightarrow P(\overline E) = \dfrac{125-1}{125}\\\Rightarrow P(\overline E) = \dfrac{124}{125}[/tex]
So, the answers are:
(a) 125
[tex](b) \dfrac{1}{125}[/tex]
[tex](c) \dfrac{124}{125}[/tex]
Angie and Brian are running a bakery. They are going to make three kinds of banana bread: regular, one with extra sugar, and one with extra bananas. Each loaf of regular takes 3 bananas, 3 cups of sugar, and 2 cups of flour. Each loaf of extra sugar takes 2 bananas, 4 cups of sugar, and 3 cups of flour. Each loaf of extra bananas takes 5 bananas, 1 cup of sugar and 4 cups of flour. They have 1200 bananas, 1050 cups of sugar, and 1075 cups of flour available. How many loaves of each type should they make to use up all the bananas, sugar and flour available
Answer:
150 regular loaves
125 loaves with extra sugar
100 loaves with extra bananas.
Step-by-step explanation:
Let R represent loaves of regular bread, S represent loaves with extra sugar, and B represent loaves with extra bananas. If they have 1200 bananas, 1050 cups of sugar, and 1075 cups of flour available, the number of each type of bread that Angie and Brian can make are given by the following system of equations:
[tex]3R+2S+5B=1,200\\3R+4S+1B=1,050\\2R+3S+4B=1,075[/tex]
Solving the linear system:
[tex]3R+4S+1B-(3R+2S+5B)=1,050-1,200\\2S-4B=-150\\S=-75+2B\\\\3R+2*(-75+2B)+5B=1,200\\2R+3*(-75+2B)+4B=1,075\\3R+9B=1,350\\2R+10B=1,300\\\\3R+9B-1.5*(2R+10B)=1,350-(1.5*1,300)\\9B-15B=-600\\B=100\\2R+10*100=1,300\\R=150\\3*150+2S+5*100=1,200\\S=125[/tex]
They should make 150 regular loaves, 125 loaves with extra sugar, and 100 loaves with extra bananas.
5/a - 4/b as a single fraction
Answer:
I'm not completely sure what you mean by a, "single fraction," but I'm pretty sure the answer you are looking for is [tex]\frac{5-4}{a-b}[/tex]
Step-by-step explanation:
The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.5 ppm and standard deviation 1.4 ppm. 18 randomly selected large cities are studied. Round all answers to two decimal places.
A. xBar~ N( ____) (____)
B. For the 18 cities, find the probability that the average amount of pollutants is more than 9 ppm.
C. What is the probability that one randomly selected city's waterway will have more than 9 ppm pollutants?
D. Find the IQR for the average of 18 cities.Q1 =
Q3 =
IQR:
2. X ~ N(30,10). Suppose that you form random samples with sample size 4 from this distribution. Let xBar be the random variable of averages. Let ΣX be the random variable of sums. Round all answers to two decimal places.
A. xBar~ N(___) (____)
B. P(xBar<30) =
C. Find the 95th percentile for the xBar distribution.
D. P(xBar > 36)=
E. Q3 for the xBar distribution =
Answer:
1)
A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)
B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552
C) P(X> 9)= 0.36317
D) IQR= 0.4422
2)
A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)
B) P( [tex]\frac{}{X}[/tex]<30)= 0.50
C) P₉₅= 32.60
D) P( [tex]\frac{}{X}[/tex]>36)= 0
E) Q₃: 31.0586
Step-by-step explanation:
Hello!
1)
The variable of interest is
X: pollutants found in waterways near a large city. (ppm)
This variable has a normal distribution:
X~N(μ;σ²)
μ= 8.5 ppm
σ= 1.4 ppm
A sample of 18 large cities were studied.
A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:
[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)
The population mean is the same as the mean of the variable
μ= 8.5 ppm
The standard deviation is
σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108
So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)
B)
P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)
To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.
Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]
Then using the Z table you'll find the probability of
P(Z≤1.51)= 0.93448
Then
1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552
C)
In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:
P(X> 9)= 1 - P(X ≤ 9)
Z= (X-μ)/δ= (9-8.5)/1.44
Z= 0.347= 0.35
P(Z≤0.35)= 0.63683
Then
P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317
D)
The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:
Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25
Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:
P(Z≤z₁)= 0.25
Using the table you have to identify the value of Z that accumulates 0.25 of probability:
z₁= -0.67
Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:
z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)
z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)
[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5= 8.2789 ppm
The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:
Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75
⇒ P(Z≤z₃)= 0.75
Using the table you have to identify the value of Z that accumulates 0.75 of probability:
z₃= 0.67
Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:
z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)
z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)
[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5= 8.7211 ppm
IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422
2)
A)
X ~ N(30,10)
For n=4
[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)
Population mean μ= 30
Population variance σ²/n= 10/4= 2.5
Population standard deviation σ/√n= √2.5= 1.58
[tex]\frac{}{X}[/tex] ~ N(30;2.5)
B)
P( [tex]\frac{}{X}[/tex]<30)
First you have to standardize the value and then look for the probability:
Z= ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0
P(Z<0)= 0.50
Then
P( [tex]\frac{}{X}[/tex]<30)= 0.50
Which is no surprise since 30 y the value of the mean of the distribution.
C)
P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95
P( Z≤ z₀)= 0.95
z₀= 1.645
Now you have to reverse the standardization:
z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)
z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)
[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60
P₉₅= 32.60
D)
P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0
E)
Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75
⇒ P(Z≤z₃)= 0.75
z₃= 0.67
z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)
z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)
[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ
[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586
Q₃: 31.0586
Simplify the expression by combining like terms
15 + 12x – 5.2 + 4y - 7
9.8 +12x+y-7
2.8+12x+4y
f(x) =3x+4=g(x)=2x+1 /x_4:find the (fog)?
Answer:
Step-by-step explanation:
on monday, it took 3 builders 5 1/2 hours to build a wall. an identical wall needs to be built on tuesday and 5 builders are available. each builder is paid £8.90 for each hour they work. work out how much each builder will be paid for the work completed on tuesday
Answer:
£29.37
Step-by-step explanation:
→ First step is to find the amount of hours it takes for 5 builders
[tex]\frac{3*\frac{11}{2} }{5} =\frac{33}{2} /5=\frac{33}{2} *\frac{1}{5} =\frac{33}{10} =3\frac{3}{10}[/tex]
→ Now we know how long 5 builder takes we need to multiply the hourly rate by their time worked
[tex]3\frac{3}{10} *8.90=\frac{33}{10} *8.90=3.3*8.90 = 29.37[/tex]
Answer:
Step-by-step explanation:
When the number of builders is increased, the hours worked will be reduced.
So, this is inverse proportion.
Number of hours worked by 5 builders = [tex]\frac{3*\frac{11}{2}}{5}\\\\[/tex]
[tex]=3*\frac{11}{2}*\frac{1}{5}\\\\=\frac{33}{10}\\\\=3\frac{1}{10}[/tex]
Amount received by each builder= 33/10 * 8.90
= £ 29.37
What is the relative change from 6546 to 4392
Answer:
The relative change from 6546 and 4392 is 49.04
Step-by-step explanation:
An airport is located next to a housing development. Profits to the airport are simply 20 f-f 2, where f is the number of flights per day. The housing developers profits are 28hh2-h, where h is the number of houses and f is the number of flights per day. If the airport is not required to pay the developer for any "damages" from the flights, how many houses will the developer build
Answer:
The total number of houses are "9". The further explanation is given below.
Step-by-step explanation:
The given values are:
height,
h = 28h - h²
Housing profit of developers will be:
⇒ [tex]\pi^h=28h-h^2-hf[/tex]
If airport won't pay any cost for the damage,
⇒ [tex]\pi^A=20f-f^2[/tex]
then,
⇒ [tex]\frac{\partial \pi^A}{\partial f}[/tex] = [tex]20-2f =0[/tex]
[tex]20=2f[/tex]
[tex]f=\frac{20}{2}[/tex]
[tex]f=10[/tex]
On putting the value of "f", we get
⇒ [tex]\pi^h=28h-h^2-10h[/tex]
[tex]=18h-h^2[/tex]
⇒ [tex]\frac{\partial \pi h}{\partial h}=18-2h=0[/tex]
[tex]2h=18[/tex]
[tex]h=\frac{18}{2}[/tex]
[tex]h=9[/tex]
So that the total number of house built by the developers will be "9".
SOMEONE PLEASE HELP ME I WILLG IVE EVERYTHING!
Answer:
AB is around 33.18BC is around 15.58Step-by-step explanation:
adjacent/hypotenuse is sine:
cos(28 degrees)=29.3/x
cos(28 degrees)*x=29.3
x=29.3/cos(28 degrees)
x=around 33.18
AB is around 33.18
opposite/adjacent is tangent
tan(28 degrees)=x/29.3
tan(28 degrees)*29.3=x
x=tan(28 degrees)*29.3
x=around 15.58
BC is around 15.58
A line passes through the points (–1, 10) and (3, 2). Which shows the graph of this line?
Answer:
Hope this helps!!
Answer:
the correct answer is ...... A
Step-by-step explanation:
Hope this helps
SOMEONE PLEASE HELP ME ASAP PLEASE!!!
Answer:
38 units
Step-by-step explanation:
We can find the perimeter of the shaded figure be finding out the number of unit lengths we have along the boundary of the given figure.
Thus, see attachment below for the number of units of each length of the figure that we have counted.
The perimeter of the figure = sum of all the lengths = 7 + 7 + 10 + 2 + 2 + 6 + 2 + 2 = 38
Perimeter of the shaded figure = 38 units
solve 2cos^2x+3sinx=0
Answer:
[tex] x = {\sin^{ - 1} 2} \: \: or \: \: x = \frac{7\pi}{6}, \: \: \frac{5\pi}{3} [/tex]
Step-by-step explanation:
[tex]2 { \cos}^{2} x + 3 \sin x = 0 \\ 2 {(1 - \sin}^{2} x) + 3 \sin x = 0 \\ 2 - 2 \sin^{2} x+ 3 \sin x = 0 \\ 2 \sin^{2} x - 3 \sin x - 2 = 0 \\ 2 \sin^{2} x - 4\sin x + \sin x- 2 = 0 \\ 2\sin x(\sin x - 2) + 1(\sin x - 2) = 0 \\ (\sin x - 2)(2\sin x + 1) = 0 \\ (\sin x - 2) = 0 \: or \: (2\sin x + 1) = 0 \\ \sin x = 2 \: or \: 2\sin x = - 1 \\ x = {\sin^{ - 1} 2} \: \: or \: \: \sin x = - \frac{1}{2} \\ x = {\sin^{ - 1} 2} \: \: or \: \: x = \frac{7\pi}{6}, \: \: \frac{5\pi}{3} [/tex]
